Longest Increasing Subsequence (LIS)
Last Updated :
23 Jul, 2025
Given an array arr[] of size n, the task is to find the length of the Longest Increasing Subsequence (LIS) i.e., the longest possible subsequence in which the elements of the subsequence are sorted in increasing order.
Examples:
Input: arr[] = [3, 10, 2, 1, 20]
Output: 3
Explanation: The longest increasing subsequence is 3, 10, 20
Input: arr[] = [30, 20, 10]
Output:1
Explanation: The longest increasing subsequences are [30], [20] and [10]
Input: arr[] = [2, 2, 2]
Output: 1
Explanation: We consider only strictly increasing.
Input: arr[] = [10, 20, 35, 80]
Output: 4
Explanation: The whole array is sorted
[Naive Approach] Using Recursion - Exponential Time and Linear Space
The idea to do traverse the input array from left to right and find length of the Longest Increasing Subsequence (LIS) ending with every element arr[i]. Let the length found for arr[i] be L[i]. At the end we return maximum of all L[i] values. Now to compute L[i], we use recursion, we consider all smaller elements on left of arr[i], recursively compute LIS value for all the smaller elements on left, take the maximum of all and add 1 to it. If there is no smaller element on left of an element, we return 1.
Let L(i) be the length of the LIS ending at index i such that arr[i] is the last element of the LIS. Then, L(i) can be recursively written as:
- L(i) = 1 + max(L(prev) ) where 0 < prev < i and arr[prev] < arr[i]; or
- L(i) = 1, if no such prev exists.
Formally, the length of LIS ending at index i, is 1 greater than the maximum of lengths of all LIS ending at some index prev such that arr[prev] < arr[i] where prev < i.
After we fill the L array, we find LIS as maximum of all in L[]
Overall LIS = max(L[i]) where 0 <= i < n
We can see that the above recurrence relation follows the optimal substructure property. Follow the below illustration to see overlapping subproblems.
Consider arr[] = [3, 10, 2, 11]
L(i): Denotes LIS of subarray ending at position 'i'
Recursion Tree
C++
// Cpp program to find lis using recursion
// in Exponential Time and Linear Space
#include <bits/stdc++.h>
using namespace std;
// Returns LIS of subarray ending with index i.
int lisEndingAtIdx(vector<int>& arr, int idx) {
// Base case
if (idx == 0)
return 1;
// Consider all elements on the left of i,
// recursively compute LISs ending with
// them and consider the largest
int mx = 1;
for (int prev = 0; prev < idx; prev++)
if (arr[prev] < arr[idx])
mx = max(mx, lisEndingAtIdx(arr, prev) + 1);
return mx;
}
int lis(vector<int>& arr) {
int n = arr.size();
int res = 1;
for (int i = 1; i < n; i++)
res = max(res, lisEndingAtIdx(arr, i));
return res;
}
int main() {
vector<int> arr = { 10, 22, 9, 33, 21, 50, 41, 60 };
cout << lis(arr);
return 0;
}
// Java program to find lis using recursion
// in Exponential Time and Linear Space
import java.util.*;
class GfG {
static int lisEndingAtIdx(int[] arr, int idx) {
// Base case
if (idx == 0)
return 1;
// Consider all elements on the left of i,
// recursively compute LISs ending with
// them and consider the largest
int mx = 1;
for (int prev = 0; prev < idx; prev++)
if (arr[prev] < arr[idx])
mx = Math.max(mx, lisEndingAtIdx(arr, prev) + 1);
return mx;
}
static int lis(int[] arr) {
int n = arr.length;
int res = 1;
for (int idx = 1; idx < n; idx++)
res = Math.max(res, lisEndingAtIdx(arr, idx));
return res;
}
public static void main(String[] args) {
int[] arr = { 10, 22, 9, 33, 21, 50, 41, 60 };
System.out.println(lis(arr));
}
}
# Python program to find lis using recursion
# in Exponential Time and Linear Space
def lisEndingAtIdx(arr, idx):
# Base case
if idx == 0:
return 1
# Consider all elements on the left of i,
# recursively compute LISs ending with
# them and consider the largest
mx = 1
for prev in range(idx):
if arr[prev] < arr[idx]:
mx = max(mx, lisEndingAtIdx(arr, prev) + 1)
return mx
def lis(arr):
n = len(arr)
res = 1
for idx in range(1, n):
res = max(res, lisEndingAtIdx(arr, idx))
return res
if __name__ == "__main__":
arr = [10, 22, 9, 33, 21, 50, 41, 60]
print(lis(arr))
// C# program to find lis using recursion
// in Exponential Time and Linear Space
using System;
using System.Collections.Generic;
class GfG {
static int lisEndingAtIdx(List<int> arr, int idx) {
// Base case
if (idx == 0)
return 1;
// Consider all elements on the left of i,
// recursively compute LISs ending with
// them and consider the largest
int mx = 1;
for (int prev = 0; prev < idx; prev++)
if (arr[prev] < arr[idx])
mx = Math.Max(mx, lisEndingAtIdx(arr, prev) + 1);
return mx;
}
static int lis(List<int> arr) {
int n = arr.Count;
int res = 1;
for (int idx = 1; idx < n; idx++)
res = Math.Max(res, lisEndingAtIdx(arr, idx));
return res;
}
static void Main(string[] args) {
List<int> arr = new List<int> { 10, 22, 9, 33, 21, 50, 41, 60 };
Console.WriteLine(lis(arr));
}
}
// JavaScript program to find lis using recursion
// in Exponential Time and Linear Space
function lisEndingAtIdx(arr, idx) {
// Base case
if (idx === 0)
return 1;
// Consider all elements on the left of i,
// recursively compute LISs ending with
// them and consider the largest
let mx = 1;
for (let prev = 0; prev < idx; prev++) {
if (arr[prev] < arr[idx]) {
mx = Math.max(mx,
lisEndingAtIdx(arr, prev) + 1);
}
}
return mx;
}
function lis(arr) {
let n = arr.length;
let res = 1;
for (let idx = 1; idx < n; idx++) {
res = Math.max(res, lisEndingAtIdx(arr, idx));
}
return res;
}
let arr = [ 10, 22, 9, 33, 21, 50, 41, 60 ];
console.log(lis(arr));
[Better Approach - 1] Using Memoization - O(n^2) Time and O(n) Space
If notice carefully, we can see that the above recursive function lisEndingAtIdx() also follows the overlapping subproblems property i.e., same substructure solved again and again in different recursion call paths. We can avoid this using the memoization approach. Since there is only one parameter that changes in recursive calls and the range of the parameter goes from 0 to n-1, so we us a 1D array of size n and initialize it as -1 to indicate that the values are not computed yet.
C++
#include <bits/stdc++.h>
using namespace std;
int lisEndingAtIdx(vector<int>& arr, int idx, vector<int>& memo) {
// Base case
if (idx == 0)
return 1;
// Check if the result is already computed
if (memo[idx] != -1)
return memo[idx];
// Consider all elements on left of i,
// recursively compute LISs ending with
// them and consider the largest
int mx = 1;
for (int prev = 0; prev < idx; prev++)
if (arr[prev] < arr[idx])
mx = max(mx, lisEndingAtIdx(arr, prev, memo) + 1);
// Store the result in the memo array
memo[idx] = mx;
return memo[idx];
}
int lis(vector<int>& arr) {
int n = arr.size();
vector<int> memo(n, -1);
int res = 1;
for (int i = 1; i < n; i++)
res = max(res, lisEndingAtIdx(arr, i, memo));
return res;
}
int main() {
vector<int> arr = { 10, 22, 9, 33, 21, 50, 41, 60 };
cout << lis(arr);
return 0;
}
import java.util.*;
class GfG {
static int lisEndingAtIdx(int[] arr, int idx, int[] memo) {
// Base case
if (idx == 0)
return 1;
// Check if the result is already computed
if (memo[idx] != -1)
return memo[idx];
// Consider all elements on left of i,
// recursively compute LISs ending with
// them and consider the largest
int mx = 1;
for (int prev = 0; prev < idx; prev++)
if (arr[prev] < arr[idx])
mx = Math.max(mx, lisEndingAtIdx(arr, prev, memo) + 1);
// Store the result in the memo array
memo[idx] = mx;
return memo[idx];
}
static int lis(int[] arr) {
int n = arr.length;
int[] memo = new int[n];
Arrays.fill(memo, -1);
int res = 1;
for (int idx = 1; idx < n; idx++)
res = Math.max(res, lisEndingAtIdx(arr, idx, memo));
return res;
}
public static void main(String[] args) {
int[] arr = {10, 22, 9, 33, 21, 50, 41, 60};
System.out.println(lis(arr));
}
}
def lisEndingAtIdx(arr, idx, memo):
# Base case
if idx == 0:
return 1
# Check if the result is already computed
if memo[idx] != -1:
return memo[idx]
# Consider all elements on left of i,
# recursively compute LISs ending with
# them and consider the largest
mx = 1
for prev in range(idx):
if arr[prev] < arr[idx]:
mx = max(mx, lisEndingAtIdx(arr, prev, memo) + 1)
# Store the result in the memo array
memo[idx] = mx
return memo[idx]
def lis(arr):
n = len(arr)
memo = [-1] * n
res = 1
for idx in range(1, n):
res = max(res, lisEndingAtIdx(arr, idx, memo))
return res
if __name__ == "__main__":
arr = [10, 22, 9, 33, 21, 50, 41, 60]
print(lis(arr))
using System;
using System.Collections.Generic;
class GfG {
static int lisEndingAtIdx(List<int> arr, int idx, int[] memo) {
// Base case
if (idx == 0)
return 1;
// Check if the result is already computed
if (memo[idx] != -1)
return memo[idx];
// Consider all elements on left of i,
// recursively compute LISs ending with
// them and consider the largest
int mx = 1;
for (int prev = 0; prev < idx; prev++)
if (arr[prev] < arr[idx])
mx = Math.Max(mx, lisEndingAtIdx(arr, prev, memo) + 1);
// Store the result in the memo array
memo[idx] = mx;
return memo[idx];
}
static int lis(List<int> arr) {
int n = arr.Count;
int[] memo = new int[n];
for(int i = 0; i < n; i++) {
memo[i] = -1;
}
int res = 1;
for (int idx = 1; idx < n; idx++)
res = Math.Max(res, lisEndingAtIdx(arr, idx, memo));
return res;
}
static void Main(string[] args) {
List<int> arr = new List<int> {10, 22, 9, 33, 21, 50, 41, 60};
Console.WriteLine(lis(arr));
}
}
function lisEndingAtIdx(arr, idx, memo) {
// Base case
if (idx === 0)
return 1;
// Check if the result is already computed
if (memo[idx] !== -1)
return memo[idx];
// Consider all elements on left of i,
// recursively compute LISs ending with
// them and consider the largest
let mx = 1;
for (let prev = 0; prev < idx; prev++) {
if (arr[prev] < arr[idx]) {
mx = Math.max(
mx, lisEndingAtIdx(arr, prev, memo) + 1);
}
}
// Store the result in the memo array
memo[idx] = mx;
return memo[idx];
}
function lis(arr) {
const n = arr.length;
const memo = Array(n).fill(-1);
let res = 1;
for (let idx = 1; idx < n; idx++) {
res = Math.max(res, lisEndingAtIdx(arr, idx, memo));
}
return res;
}
const arr = [ 10, 22, 9, 33, 21, 50, 41, 60 ];
console.log(lis(arr));
[Better Approach - 2] Using DP (Bottom Up Tabulation) - O(n^2) Time and O(n) Space
The tabulation approach for finding the Longest Increasing Subsequence (LIS) solves the problem iteratively in a bottom-up manner. The idea is to maintain a 1D array lis[], where lis[i] stores the length of the longest increasing subsequence that ends at index i. Initially, each element in lis[] is set to 1, as the smallest possible subsequence for any element is the element itself.
The algorithm then iterates over each element of the array. For each element arr[i], it checks all previous elements arr[0] to arr[i-1]. If arr[i] is greater than arr[prev] (ensuring the subsequence is increasing), it updates lis[i] to the maximum of its current value or lis[prev] + 1, indicating that we can extend the subsequence ending at arr[prev] by including arr[i].
Finally, the length of the longest increasing subsequence is the maximum value in the lis[] array.
C++
#include <bits/stdc++.h>
using namespace std;
// lis() returns the length of the longest
// increasing subsequence in arr of size n
int lis(vector<int>& arr) {
int n = arr.size();
vector<int> lis(n, 1);
// Compute optimized LIS values in
// bottom-up manner
for (int i = 1; i < n; i++) {
for (int prev = 0; prev < i; prev++) {
if (arr[i] > arr[prev] && lis[i] < lis[prev] + 1) {
lis[i] = lis[prev] + 1;
}
}
}
// Return maximum value in lis
return *max_element(lis.begin(), lis.end());
}
int main() {
vector<int> arr = { 10, 22, 9, 33, 21, 50, 41, 60 };
cout << lis(arr) << endl;
return 0;
}
import java.lang.*;
class GfG {
// lis() returns the length of the longest
// increasing subsequence in arr[] of size n
static int lis(int arr[]) {
int n = arr.length;
int lis[] = new int[n];
// Initialize LIS values for all indexes
for (int i = 0; i < n; i++)
lis[i] = 1;
// Compute optimized LIS values in
// bottom up manner
for (int i = 1; i < n; i++)
for (int prev = 0; prev < i; prev++)
if (arr[i] > arr[prev] && lis[i] < lis[prev] + 1)
lis[i] = lis[prev] + 1;
// Pick maximum of all LIS values
int max = 1;
for (int i = 0; i < n; i++)
max = Math.max(max, lis[i]);
return max;
}
public static void main(String args[]) {
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
System.out.println(lis(arr));
}
}
# lis returns length of the longest
# increasing subsequence in arr of size n
def lis(arr):
n = len(arr)
# Declare the list (array) for LIS and
# initialize LIS values for all indexes
lis = [1] * n
# Compute optimized LIS values in bottom
# -up manner
for i in range(1, n):
for prev in range(0, i):
if arr[i] > arr[prev]:
lis[i] = max(lis[i], lis[prev] + 1)
# Return the maximum of all LIS values
return max(lis)
if __name__ == '__main__':
arr = [10, 22, 9, 33, 21, 50, 41, 60]
print(lis(arr))
using System;
class GfG {
// lis() returns the length of the longest
// increasing subsequence in arr[] of size n
static int lis(int[] arr) {
int n = arr.Length;
int[] lis = new int[n];
// Initialize LIS values for all indexes
for (int i = 0; i < n; i++)
lis[i] = 1;
// Compute optimized LIS values in bottom up manner
for (int i = 1; i < n; i++) {
for (int prev = 0; prev < i; prev++) {
if (arr[i] > arr[prev] && lis[i] < lis[prev] + 1)
lis[i] = lis[prev] + 1;
}
}
// Pick maximum of all LIS values
int max = 0;
for (int i = 0; i < n; i++) {
if (max < lis[i])
max = lis[i];
}
return max;
}
static void Main() {
int[] arr = { 10, 22, 9, 33, 21, 50, 41, 60 };
Console.WriteLine(lis(arr));
}
}
// lis() returns the length of the longest
// increasing subsequence in arr[] of size n
function lis(arr) {
let n = arr.length;
let lis = Array(n).fill(0);
let max = 0;
// Initialize LIS values for all indexes
for (let i = 0; i < n; i++)
lis[i] = 1;
// Compute optimized LIS values in
// bottom up manner
for (let i = 1; i < n; i++) {
for (let prev = 0; prev < i; prev++) {
if (arr[i] > arr[prev] && lis[i] < lis[prev] + 1)
lis[i] = lis[prev] + 1;
}
}
// Pick maximum of all LIS values
for (let i = 0; i < n; i++)
if (max < lis[i])
max = lis[i];
return max;
}
let arr = [10, 22, 9, 33, 21, 50, 41, 60];
console.log(lis(arr));
[Expected Approach] Using Binary Search - O(n Log n) Time and O(n) Space
We can solve this in O(n Log n) time using Binary Search. The idea is to traverse the given sequence and maintain a separate list of sorted subsequence so far. For every new element, find its position in the sorted subsequence using Binary Search.
Refer Longest Increasing Subsequence Size (N * logN) for details.
Problems based on LIS
Longest Increasing Subsequence (LIS)
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